Electromagnetic Waves

JEE Main level, single correct, medium

128 of 165

Question 128 of 165

During the propagation of electromagnetic wave in a particular medium :-

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Answer: (C) Electric energy density is equal to the magnetic energy density

(UV)elec=12∈0E2{\left(\dfrac{U}{V}\right)}_{elec}=\dfrac{1}{2}{\in }_{0}{E}^{2} ....(i)

(UV)mag=12B2/μ0{\left(\dfrac{U}{V}\right)}_{mag}=\dfrac{1}{2}{B}^{2}/{\mu }_{0} ....(ii)

B=EC\text{B}=\dfrac{\text{E}}{\text{C}} (for EMW) ....(iii)

and C=1μ0ϵ0C=\dfrac{1}{\sqrt{{\mu }_{0}{\epsilon }_{0}}} .....(iv)

∴ devide eqn (i) by (ii)

(U/V)elec(U/V)mag=12ϵ0E212B2/μ0\dfrac{{(U/V)}_{elec}}{{(U/V)}_{mag}}=\dfrac{\dfrac{1}{2}{\epsilon }_{0}{E}^{2}}{\dfrac{1}{2}{B}^{2}/{\mu }_{0}}

=(EB)2μ0ϵ0={(\dfrac{E}{B})}^{2}{\mu }_{0}{\epsilon }_{0}
by using eq (iii) & (iv)

=C2C2=1=\dfrac{{C}^{2}}{{C}^{2}}=1

⇒(U/V)elec.=(UV)mag\Rightarrow {(U/V)}_{elec.}={(\dfrac{U}{V})}_{mag}

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